Question

Quantos termos tem na PG (3,6...,,384

Answer 1

An=A1×q^(n-1)

A1=3
A2=6


q=An/An-1

q=A2/A1=6/3=2

An=384 n=?

An=3×2^(n-1)
384=3×2^(n-1)

384/3=2^(n-1)
128=2^(n-1)

Decompondo o 128 em dois teremos

2^7=2^(n-1)

7=n-1
7+1=n

8=n